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Stella [2.4K]
3 years ago
10

What are the differences between simple, compound, and electron microscopes? WILL GET BRAINIEST!

Chemistry
1 answer:
timurjin [86]3 years ago
5 0
I think the answer is= simple/uses surrounding light source and is restricted in magnification.
compound uses an electrical light source but is restricted in magnification also. an electron microscope has electrical magnification and light source so you can see smaller cells when dyed the rt. color..plse double check, good luck..
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If 56.8 ml of bacl2 solution is needed to precipitate all the sulfate ion in a 554 mg sample of na2so4 (forming baso4), what is
IrinaVladis [17]
<span>0.0687 m The balanced equation is BaCl2 + Na2SO4 ==> BaSO4 + 2 NaCl Looking at the equation, it indicates that there's a 1 to 1 ratio of BaCl2 and Na2SO4 in the reaction. So the number of moles of each will be equal. Now calculate the number of moles of Na2SO4 we had. Start by looking up atomic weights. Atomic weight sodium = 22.989769 Atomic weight sulfur = 32.065 Atomic weight oxygen = 15.999 Molar mass Na2SO4 = 2 * 22.989769 + 32.065 + 4 * 15.999 = 142.040538 g/mol Moles Na2SO4 = 0.554 g / 142.040538 g/mol = 0.003900295 mol Molarity is defined as moles per liter, so let's do the division. 0.003900295 mol / 0.0568 l = 0.068667165 mol/l = 0.068667165 m Rounding to 3 significant figures gives 0.0687 m</span>
3 0
3 years ago
Someone help me i don’t understand it
Troyanec [42]
Use PV=nRT to help you
3 0
3 years ago
Balance the following redox reaction in acidic solution. Zn(s)+MnO−4(aq)→ Zn+2(aq)+Mn+2(aq)
devlian [24]

Answer:

2MnO₄⁻ + 5Zn + 16H⁺ → 2Mn²⁺ + 8H₂O + 5Zn²⁺

Explanation:

To balance a redox reaction in an acidic medium, we simply follow some rules:

  1. Split the reaction into an oxidation and reduction half.
  2. By inspecting, balance the half equations with respect to the charges and atoms.
  3. In acidic medium, one atom of H₂O is used to balance up each oxygen atom and one H⁺ balances up each hydrogen atom on the deficient side of the equation.
  4. Use electrons to balance the charges. Add the appropriate numbers of electrons the side with more charge and obtain a uniform charge on both sides.
  5. Multiply both equations with appropriate factors to balance the electrons in the two half equations.
  6. Add up the balanced half equations and cancel out any specie that occur on both sides.
  7. Check to see if the charge and atoms are balanced.

Solution

                            Zn + MnO₄⁻ → Zn²⁺ + Mn²⁺

The half equations:

                      Zn → Zn²⁺                          Oxidation half

                      MnO₄⁻ → Mn²⁺                  Reduction half

Balancing of atoms(in acidic medium)

                     Zn → Zn²⁺

                    MnO₄⁻ + 8H⁺ → Mn²⁺ + 4H₂O

Balancing of charge

                   Zn → Zn²⁺ + 2e⁻

                    MnO₄⁻ + 8H⁺ + 5e⁻→ Mn²⁺ + 4H₂O

Balancing of electrons

         Multiply the oxidation half by 5 and reduction half by 2:

                          5Zn → 5Zn²⁺ + 10e⁻

                        2MnO₄⁻ + 16H⁺ + 10e⁻→ 2Mn²⁺ + 8H₂O

Adding up the two equations gives:

              5Zn + 2MnO₄⁻ + 16H⁺ + 10e⁻ → 5Zn²⁺ + 10e⁻ + 2Mn²⁺ + 8H₂O

The net equation gives:

         5Zn + 2MnO₄⁻ + 16H⁺ → 5Zn²⁺ + 2Mn²⁺ + 8H₂O

8 0
3 years ago
X: glucose + oxygen ---- carbon dioxide + water
elena-s [515]

Answer:

X

Explanation:

5 0
2 years ago
Read 2 more answers
After decaying for 48 hours, one-sixteenth (1/16) of the original mass of a radioisotope sample remains unchanged. What is the h
Hunter-Best [27]

The half-life of this radioisotope : 12 hr

<h3>Further explanation </h3>

The atomic nucleus can experience decay into 2 particles or more due to the instability of its atomic nucleus.  

Usually radioactive elements have an unstable atomic nucleus.  

General formulas used in decay:  

\large{\boxed{\bold{N_t=N_0(\dfrac{1}{2})^{t/t\frac{1}{2} }}}

t = duration of decay  

t 1/2 = half-life  

N₀ = the number of initial radioactive atoms  

Nt = the number of radioactive atoms left after decaying during T time  

t=48 hr

\tt \dfrac{Nt}{No}=\dfrac{1}{16}

The half-life :

\tt \dfrac{1}{16}=\dfrac{1}{2}^{(48/t\frac{1}{2} )}\\\\(\dfrac{1}{2})^4=(\dfrac{1}{2})^{48/t\frac{1}{2}}\\\\4=48/t\frac{1}{2}\\\\t\frac{1}{2}=12~hr

7 0
3 years ago
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